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Which test?Determine which chi-square test is appropriate in each of the following settings. Explain your reasoning.

a. Does chocolate help heart-attack victims live longer? Researchers in Sweden randomly selected 1169people who had suffered heart attacks and asked them about their consumption of chocolate in the previous year. Then the researchers followed these people and recorded whether or not they had died within 8years.

b. Random-digit-dialing telephone surveys used to exclude cell-phone numbers. If the opinions of people who have only cell phones differ from those of people who have landline service, the poll results may not represent the entire adult population. The Pew Research Center interviewed separate random samples of cell-only and landline telephone users who were less than 30years old and asked them to describe their political party affiliation

Short Answer

Expert verified

a. We should use Chi-square test for independence.

b. We should use Chi-square test for homogeneity.

Step by step solution

01

Part (a) Step 1 : Given information

We have to determine which chi-square test is appropriate for given setting.

02

Part (a) Step 2 : Simplification

Does chocolate help those who have had a heart attack live longer?
Researchers in Sweden chose 1169persons who had heart attacks at random and asked them about their chocolate consumption in the previous year. The researchers then followed up with these folks to see if they died within the next eight years.
First, we must determine which test should be used in a certain case. There are three tests to complete:
Chi-square goodness-of-fit test, chi-square homogeneity test, and chi-square independence test All of these tests will be detailed for us. A chi-square goodness-of-fit test is used when we are interested in the distribution of a single variable. In this circumstance, we must employ a chi-square test for homogeneity when we are interested in the distribution of two variables with numerous independent samples. We would like to do a chi-square test for independence when we are interested in the distribution of two variables and there is only one sample. We are provided two variables in the present setting: chocolate consumption and death after 8years.
One random sample of 1169patients who experienced heart attacks was taken.
As a result, we should perform the Chi-square test to determine independence.
03

Part (b) Step 1 : Given information

We have to determine which chi-square test is appropriate for given setting.

04

Part (b) Step 2 : Simplification

Cell phone numbers were previously excluded from random-digit-dialing telephone polls. The poll findings may not represent the full adult population if the opinions of persons who only have cell phones differ from those of people who have landline service. The Pew Research Center questioned separate random samples of cell-only and landline phone users under the age of 30to ask them about their political party allegiance.
First, we must determine which test should be used in a certain case. There are three tests to complete:
Chi-square goodness-of-fit test, chi-square homogeneity test, and chi-square independence test
All of these tests will be detailed for us. A chi-square goodness-of-fit test is used when we are interested in the distribution of a single variable. In this circumstance, we must employ a chi-square test for homogeneity when we are interested in the distribution of two variables with numerous independent samples. We would like to do a chi-square test for independence when we are interested in the distribution of two variables and there is only one sample. There are two variables in this case. Telephone model and political party There are two separate random samples of cell phone and landline users.
Asaresult,weshouldperformtheChi-squaretesttodeterminehomogeneity.

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Most popular questions from this chapter

You say tomato The paper “Linkage Studies of the Tomato” (Transactions of the Canadian Institute, 1931) reported the following data on phenotypes resulting from crossing tall cut-leaf tomatoes with dwarf potato-leaf tomatoes. We wish to investigate whether the following frequencies are consistent with genetic laws, which state that the phenotypes should occur in the ratio 9:3:3:1

Assume that the conditions for inference are met. Carry out a test at the α=0.05 significance level of the proposed genetic model.

No chi-square A school’s principal wants to know if students spend about the same amount of time on homework each night of the week. She asks a random sample of 50 students to keep track of their homework time for a week. The following table displays the average amount of time (in minutes) students reported per night.

Explain carefully why it would not be appropriate to perform a chi-square test for goodness of fit using these data.

Preventing strokes Aspirin prevents blood from clotting and so helps prevent strokes.

The Second European Stroke Prevention Study asked whether adding another anticlotting

drug named dipyridamole would be more effective for patients who had already had a

stroke. Here are the data on strokes during the two years of the study

a. Summarize these data in a two-way table.

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treatments at theα=0.05significance level?

Which of the following statements about chi-square distributions are true?

I. For all chi-square distributions, P(x20)=1

II. A chi-square distribution with fewer than 10degrees of freedom is roughly symmetric.

III. The more degrees of freedom a chi-square distribution has, the larger the mean of the distribution.

a. I only

b. II only

c. III only

d. I and III

e. I, II, and III

“Will changing the rating scale on a survey affect how people answer the question?” To find out, the group took an SRS of 50students from an alphabetical roster of the school’s just over 1000students. The first 22students chosen were asked to rate the cafeteria food on a scale of 1(terrible) to 5(excellent). The remaining 28students were asked to rate the cafeteria food on a scale of 0(terrible) to 4(excellent). Here are the data:

The students decided to compare the average ratings of the cafeteria food on the two scales.

a. Find the mean and standard deviation of the ratings for the students who were given the 1to5scale.

b. For the students who were given the 0to4scale, the ratings have a mean of 3.21and a standard deviation of 0.568. Since the scales differ by one point, the group decided to add 1to each of these ratings. What are the mean and standard deviation of the adjusted ratings?

c. Would it be appropriate to compare the means from parts (a) and (b) using a two-sample t test? Justify your answer

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